Originally Posted by Canazes9

A projectile launched at a horizontal target with an upward trajectory from the line of bore relative to the line of sight absolutely does NOT begin falling the instant it leaves the bore. It begins falling the instant it leaves barrel relative to the line of bore (again frame of reference). But relative to the only things that you can actually see when squeeze the trigger - the line of sight and terra firma, the bullet does indeed rise for a brief period of time.

No problem here. I should have said "accelerating downward". The statement I made is true in the case of a horizontal barrel.

Originally Posted by Canazes9

Being subject to the force of gravity does not equal falling. I am subject to the force of gravity right now - I'm not falling. When I get in my plane next week for my scheduled trip, I will be subject to the force of gravity, but I will in fact rise as I begin that journey (for a significant period of time I hope!) Because the bullet is no longer accelerating when it leaves the muzzle, the upward force counteracting gravity is quickly overwhelmed, and the force of gravity accelerates it's return to earth. But the energy of the bullet imparted to an upward trajectory does in fact counteract the force of gravity for a brief period of time.

Pretend I've got a target paper (no backstop) placed at 1000 yards exactly 4 feet off the ground. I fire a rifle perfectly sighted in to be dead on at 1000 yards from that rifle that has the end of the muzzle precisely 4 feet off the ground while simultaneously dropping a bullet from exactly 4' in elevation immediately next to the muzzle - which bullet hits the ground first?

Please don't tell me that you believe they will hit at the exact same time, because they won't. The bullet dropped from the hand will hit significantly sooner than the bullet fired at the 1000 yard target. The difference in impact times can only be because a force (a portion of the bullet's initial velocity) was utilized to counteract the force of gravity (briefly).

David,

With all due respect, that's not how physics works. The direction of acceleration is a function of the net force on an object, not just a single, specific force. I specifically said that the bullet begins falling once it is no longer supported by the barrel because at that point gravity is the only force acting on the bullet. The reason that you are not falling right now is that gravity is not the only force acting on your body. The normal force from the chair or the floor is exactly equal and opposite the force of gravity. But I can see why the word "falling" causes confusion. As you said, the bullet does begin falling (ie, it has a downward velocity vector) from the axis of the bore the moment it leaves the barrel, but not necessarily relative to the axis of the LOS. Perhaps a more precise statement would be that the bullet begins accelerating downward towards the earth the moment the force of gravity is the only force acting on it along the vertical axis (aka when it leaves the muzzle). This is true relative to both the LOS and the axis of the bore.

When the bullet leaves the muzzle, there is no upward force at play (assuming no aerodynamic jump, etc) to counteract gravity. There is also no energy imparted to an upward trajectory. I think what you are trying to describe is the difference between velocity and acceleration. Assuming a positive angle between the axis of the bore and the surface of the earth, as you've described, when the bullet leaves the muzzle it has an initial velocity. We can break this velocity into two vector components, a vertical component and a horizontal component. We are only concerned with the vertical component in this discussion. The bullet has an initial velocity with a positive/upward vertical component. The net force on the bullet is simply the force of gravity, which results in an immediate downward acceleration as soon as the bullet leave the muzzle. This downward acceleration immediately begins to decrease the magnitude of the upward velocity, and eventually the magnitude of the vertical velocity vector reaches zero and then becomes a downward velocity vector. This is the exact same principle as if you were to throw a ball straight up in the air, and then catch it again. The ball starts with an upward velocity, but begins accelerating downward the moment it leaves your hand. It eventually has a velocity magnitude of zero (the turning point) and starts to fall back down toward your hand.

In your example above, the bullet fired at the 1000 yard target hits the ground after the bullet dropped from the hand, not because the bullet fired from the rifle has some force acting on it as a part of its initial velocity, but because it is given an initial velocity with a positive vertical component, while the bullet dropped from the hand is not. It's just like comparing the drop times of a ball that you throw straight up and allow to hit the ground, versus one that you drop straight down from your hand and allow to hit the floor. It's obvious that the ball thrown upward will hit the ground after the ball dropped from your hand directly.

Originally Posted by Canazes9

If it helps you or anyone else purely to describe ballistics from the reference point of line of bore, I have no issue with it. To jump up and down and tell someone that bullets don't rise when people stating such are clearly referencing light of sight is silly and disingenuous.

David


I can readily describe the ballistics of a bullet from whichever reference frame you like, so it doesn't matter to me, but the reason that guys are quick to correct the statement that "bullets rise when they leave the bore" is that many people are confused by such statements, and believe that the bullet rises relative to the axis of the bore after leaving the muzzle. This is a myth that has been perpetuated for decades. Clearly you are not under that misconception, but many are.